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Let $F_1,F_2,\dots$ be sets in some sigma algebra, let $E_i = F_i \setminus\bigcup_{j<i} F_j$. It is true that now $E_i$ and $F_i$ are disjoint, but is it true that $\bigcup_{i=1}^n E_i = \bigcup_{i=1}^n F_i$? the infinite case I can see, but not the finite case

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  • $\begingroup$ What's true is that the $E_i$'s are disjoint. Also, I wonder how you can see the infinite case, because it is essentially the same proof for the finite case $\endgroup$ – user258700 May 1 '16 at 14:25
  • $\begingroup$ Normally $\text{“}{\cup}\text{''}$ is use for things like $A\cup B$ and $A_1\cup\cdots\cup A_n$, and $\text{“}\bigcup\text{''}$ for things like $\bigcup_{i=1}^n A_i$. I changed it. $\qquad$ $\endgroup$ – Michael Hardy May 1 '16 at 14:28
  • $\begingroup$ @AhmedHussein I thought $\bigcup_{i=1}^nE_i = F_n$ that's how. $\endgroup$ – coke May 1 '16 at 14:31
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$\text{“}E_i$ and $F_i$ are disjoint$\text{''}$ could be construed to mean $E_i\cap F_i=\varnothing$, and that is not true. It is true that $E_1,E_2,E_3,\ldots$ are pairwise disjoint.

Suppose $x\in\bigcup_i F_i$. Then there is some smallest index $i_0$ such that $x\in F_i$. For that smallest index $i_0$ we have $x\in E_{i_0}$; therefore $x\in\bigcup_i E_i$.

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